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Setler79 [48]
2 years ago
9

What will happen to the period of a ball-on-a-spring if the ball is replaced by a smaller one with half the mass

Physics
1 answer:
Murrr4er [49]2 years ago
4 0

When the ball on a spring is replaced by a smaller ball with half the mass :  The period decrease by 29%

<h3>Calculate the Period </h3>

We will apply the formula below to calculate period

T = 2\pi \sqrt{\frac{m}{k } }

 where :

T₁ / T₂ = \sqrt{\frac{m_{1} }{m_{2} } }    where m₂ = m₁ / 2

Therefore :

T₂ = \frac{T_{1} }{\sqrt{2} }   = 0.71 T

Hence we can conclude that the period will decrease by 29% ( i.e 100 - 71 ).

Learn more about Period calculation : brainly.com/question/10728818

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What frequency is received by a person watching an oncoming ambulance moving at 110 km/h and emitting a steady 800-Hz sound from
Bas_tet [7]

Answer:

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Explanation:

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Which type of light wave has the highest energy? which has the least amount of energy
devlian [24]

Answer:

Red has the lowest energy and violet the highest. Beyond red and violet are many other kinds of light our human eyes can't see, much like there are sounds our ears can't hear. On one end of the electromagnetic spectrum are radio waves, which have wavelengths billions of times longer than those of visible light.

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2 years ago
Why does a projectile fired along a horizontal not follow a straight path?​
Stella [2.4K]

Answer: A projectile which is fired horizontally is being constantly acted upon by acceleration due to gravity, acting vertically downwards. Hence, it does not follow a straight line path. Also Why a projectile fixed along the horizontal not follow a straight line path? Because the projectile fired horizontally is constantly acts upon by acceleration due to gravity acting vertically downwards.

Explanation:

Hope this helped :)

3 0
3 years ago
HELP PLEASE!!! 30+ points!!
GarryVolchara [31]

1) 9.26 cm

Explanation:

The focal length of a plane mirror is virtually infinite. Considering the lens equation,

\frac{1}{f}=\frac{1}{p}+\frac{1}{q}

where f is the focal length, p is the object distance, q the image distance. If we replace f with infinity, we get

q=-p

The magnification equation states that

y' = -\frac{q}{p}y

where y is the size of the object and y' the size of the image. Substituting q=-p, we get

y'=y

this means that the image produced by a plane mirror is always:

- Upright (y' is positive)

- The same size as the object

In this case, we have a book of height 9.26 cm (y=9.26 cm). This means that the magnitude of the size of the image (y') will be 9.26 cm as well.

2) 22.7 cm

As we said before, due to the infinite focal length of a plane mirror,

q=-p

this means that the image produced by a plane mirror is always:

- Virtual (because q is negative)

- At the same distance from the mirror as the object

In this case, we have a book placed at 22.7 cm from the mirror (p=22.7 cm). This means that the magnitude of the distance of the image from the mirror (q) will be 22.7 cm as well.

3) 1.60 m/s

We said previously that the image produced by a plane mirror is always at the same distance from the mirror as the real object. This implies that whenever we move the object toward/away from the mirror, the distance p will alway remain equal to the distance q. But this also means that the object and the distance are moving toward/away from the mirror at the same speed.

Therefore, since in this case the person is moving away from the mirror at 1.60 m/s, the image will also move away at a speed of 1.60 m/s.

4 0
3 years ago
The cross section of a copper strip is 1.2 mmthick and 20 mm wide. There is a 25-A current through this cross section, with the
Naily [24]

To solve this problem it is necessary to use the concepts related to the Hall Effect and Drift velocity, that is, at the speed that an electron reaches due to a magnetic field.

The drift velocity is given by the equation:

V_d = \frac{I}{nAq}

Where

I = current

n = Number of free electrons

A = Cross-Section Area

q = charge of proton

Our values are given by,

I = 25 A

A= 1.2*20 *10^{-6} m^2

q= 1.6*10^{-19}C

N = 8.47*10^{19} mm^{-3}

V_d =\frac{25}{(1.2*20 *10^{-6})(1.6*10^{-19})(8.47*10^{19} )}

V_d = 7.68*10^{-5}m/s

The hall voltage is given by

V=\frac{IB}{ned}

Where

B= Magnetic field

n = number of free electrons

d = distance

e = charge of electron

Then using the formula and replacing,

V=\frac{(2.5)(25)}{(8.47*10^{28})(1.6*10^{-19})(1.2*10^{-3})}

V = 3.84*10^{-6}V

5 0
3 years ago
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